Answer:
(a) The probability mass function of <em>X</em> is:

(b) The most likely value for <em>X</em> is 1.32.
(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number among the four homeowners who have earthquake insurance.
The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.
The random sample of homeowners selected is, <em>n</em> = 4.
The event of a homeowner having an earthquake insurance is independent of the other three homeowners.
(a)
All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.
The probability mass function of <em>X</em> is:

(b)
The most likely value of a random variable is the expected value.
The expected value of a Binomial random variable is:

Compute the expected value of <em>X</em> as follows:


Thus, the most likely value for <em>X</em> is 1.32.
(c)
Compute the probability that at least two of the four selected have earthquake insurance as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.